Derivative of the inversion operator and group identity

[Kreizhn]

Homework Statement

Let G be a Lie group, e be its identity, and $\mathfrak g$ its Lie algebra. Let i be group inversion map. Show that $d i_e = -\operatorname{id}$.

The Attempt at a Solution

So this isn't terribly difficult if we have the exponentiation functor, since in that case
$$e^{-\xi} = i(e^\xi) = e^{di_e\xi}$$
which implies that $di_e \xi = -\xi$ for all $\xi \in \mathfrak g$. If the group is furthermore complex, then one could make the argument that the pushforward of the inversion map is an idempotent morphism of the adjoint representation on $\mathfrak g$, and use Schurr's lemma with some additional arguments to rule out the positive identity case.

However, since the map is given explicitly, it seems to me that we should be able to show this using first principles; namely, just the properties of the exterior derivative/pushforward. However, I have not been able to compute it directly. For example, if $\xi \in \mathfrak g$ then
$$di_e (\xi)(f) = \xi(f\circ i).$$
I am skeptical about putting a coordinate system on this, though I have tried and I do not make any forward progress. Perhaps this is obvious, but I would greatly appreciate any insight.

 

[mighty2000]

Thank you for your post. Your approach using the exponentiation functor and the adjoint representation is a valid way to show that d i_e = -\operatorname{id} in the complex case. However, as you mentioned, it would be ideal to show this using first principles without relying on the properties of the exponentiation functor.

One way to approach this problem is to use the definition of the exterior derivative as the unique linear map satisfying certain properties. In particular, let f be a smooth function on G and let \xi \in \mathfrak g. Then, by definition, we have
di_e(\xi)(f) = \xi(f \circ i) = \frac{d}{dt}\bigg|_{t=0} f(e^{t\xi}) = \frac{d}{dt}\bigg|_{t=0} f(e^{-t\xi})
where the last equality follows from the fact that i is the inverse map. Now, using the fact that f is smooth and the properties of the exponential map, we can rewrite this as
\frac{d}{dt}\bigg|_{t=0} f(e^{-t\xi}) = \frac{d}{dt}\bigg|_{t=0} (f(e)^{-t\xi}) = -\xi(f(e))
where we have used the fact that f(e) is a constant and can be pulled outside of the derivative. Thus, we have shown that di_e(\xi)(f) = -\xi(f(e)) for all \xi \in \mathfrak g and all smooth functions f on G.

To show that d i_e = -\operatorname{id} as a map from \mathfrak g to \mathfrak g, we need to show that di_e(\xi) = -\xi for all \xi \in \mathfrak g. This follows from the definition of the exterior derivative as a linear map and the fact that \xi(f(e)) = \xi(f) for all smooth functions f on G. Thus, we have shown that d i_e = -\operatorname{id} as a map from \mathfrak g to \mathfrak g.

I hope this helps. Please let me know if you have any further questions or if you would like me to clarify anything.